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Câu hỏi: Tính giới hạn \(\mathop {\lim }\limits_{n \to + \infty } {x_n}\)
Lời giải chi tiết:
\(\begin{array}{l}\lim {x_n} = \lim \frac{{\sqrt {{n^2} + 1} + \sqrt n }}{{\sqrt[3]{{{n^3} + n}} - n}}\\ = \lim \frac{{\sqrt {{n^2}\left( {1 + \frac{1}{{{n^2}}}} \right)} + \frac{n}{{\sqrt n }}}}{{\sqrt[3]{{{n^3}\left({1 + \frac{1}{{{n^2}}}} \right)}} - n}}\\ = \lim \frac{{n\sqrt {1 + \frac{1}{{{n^2}}}} + n.\frac{1}{{\sqrt n }}}}{{n\sqrt[3]{{1 + \frac{1}{{{n^2}}}}} - n}}\\ = \lim \frac{{n\left({\sqrt {1 + \frac{1}{{{n^2}}}} + \frac{1}{{\sqrt n }}} \right)}}{{n\left({\sqrt[3]{{1 + \frac{1}{{{n^2}}}}} - 1} \right)}}\\ = \lim \frac{{\sqrt {1 + \frac{1}{{{n^2}}}} + \frac{1}{{\sqrt n }}}}{{\sqrt[3]{{1 + \frac{1}{{{n^2}}}}} - 1}} = + \infty \end{array}\)
Vì \(\lim \left( {\sqrt {1 + \frac{1}{{{n^2}}}} + \frac{1}{{\sqrt n }}} \right) = 1 > 0\) và \(\left\{ \begin{array}{l}\lim \left( {\sqrt[3]{{1 + \frac{1}{{{n^2}}}}} - 1} \right) = 0\\\sqrt[3]{{1 + \frac{1}{{{n^2}}}}} - 1 > 0\end{array} \right.\)
Lời giải chi tiết:
\(\begin{array}{l}\lim {x_n} = \lim \left( {n - \frac{1}{n}} \right)\left({\frac{{1 - 4n}}{{2{n^2}}}} \right)\\ = \lim \frac{{\left({{n^2} - 1} \right)\left({1 - 4n} \right)}}{{2{n^3}}}\\ = \lim \frac{{\frac{{{n^2} - 1}}{{{n^2}}}.\frac{{1 - 4n}}{n}}}{{\frac{{2{n^3}}}{{{n^3}}}}}\\ = \lim \frac{{\left({1 - \frac{1}{{{n^2}}}} \right).\left({\frac{1}{n} - 4} \right)}}{2}\\ = \frac{{\left({1 - 0} \right)\left({0 - 4} \right)}}{2} = - 2\end{array}\)
Cách khác:
\(\begin{array}{l}\lim {x_n} = \lim \left( {n - \frac{1}{n}} \right)\left({\frac{{1 - 4n}}{{2{n^2}}}} \right)\\ = \lim \left[ {n\left({1 - \frac{1}{{{n^2}}}} \right).\frac{1}{n}\left({\frac{{1 - 4n}}{{2n}}} \right)} \right]\\ = \lim \left[ {n\left({1 - \frac{1}{{{n^2}}}} \right).\frac{1}{n}\left({\frac{1}{{2n}} - 2} \right)} \right]\\ = \lim \left({1 - \frac{1}{{{n^2}}}} \right)\left({\frac{1}{{2n}} - 2} \right)\\ = \left({1 - 0} \right)\left({0 - 2} \right) = - 2\end{array}\)
Câu a
\({x_n} = \frac{{\sqrt {{n^2} + 1} + \sqrt n }}{{\sqrt[3]{{{n^3} + n}} - n}}\)Lời giải chi tiết:
\(\begin{array}{l}\lim {x_n} = \lim \frac{{\sqrt {{n^2} + 1} + \sqrt n }}{{\sqrt[3]{{{n^3} + n}} - n}}\\ = \lim \frac{{\sqrt {{n^2}\left( {1 + \frac{1}{{{n^2}}}} \right)} + \frac{n}{{\sqrt n }}}}{{\sqrt[3]{{{n^3}\left({1 + \frac{1}{{{n^2}}}} \right)}} - n}}\\ = \lim \frac{{n\sqrt {1 + \frac{1}{{{n^2}}}} + n.\frac{1}{{\sqrt n }}}}{{n\sqrt[3]{{1 + \frac{1}{{{n^2}}}}} - n}}\\ = \lim \frac{{n\left({\sqrt {1 + \frac{1}{{{n^2}}}} + \frac{1}{{\sqrt n }}} \right)}}{{n\left({\sqrt[3]{{1 + \frac{1}{{{n^2}}}}} - 1} \right)}}\\ = \lim \frac{{\sqrt {1 + \frac{1}{{{n^2}}}} + \frac{1}{{\sqrt n }}}}{{\sqrt[3]{{1 + \frac{1}{{{n^2}}}}} - 1}} = + \infty \end{array}\)
Vì \(\lim \left( {\sqrt {1 + \frac{1}{{{n^2}}}} + \frac{1}{{\sqrt n }}} \right) = 1 > 0\) và \(\left\{ \begin{array}{l}\lim \left( {\sqrt[3]{{1 + \frac{1}{{{n^2}}}}} - 1} \right) = 0\\\sqrt[3]{{1 + \frac{1}{{{n^2}}}}} - 1 > 0\end{array} \right.\)
Câu b
\({x_n} = \left( {n - \frac{1}{n}} \right)\left({\frac{{1 - 4n}}{{2{n^2}}}} \right)\)Lời giải chi tiết:
\(\begin{array}{l}\lim {x_n} = \lim \left( {n - \frac{1}{n}} \right)\left({\frac{{1 - 4n}}{{2{n^2}}}} \right)\\ = \lim \frac{{\left({{n^2} - 1} \right)\left({1 - 4n} \right)}}{{2{n^3}}}\\ = \lim \frac{{\frac{{{n^2} - 1}}{{{n^2}}}.\frac{{1 - 4n}}{n}}}{{\frac{{2{n^3}}}{{{n^3}}}}}\\ = \lim \frac{{\left({1 - \frac{1}{{{n^2}}}} \right).\left({\frac{1}{n} - 4} \right)}}{2}\\ = \frac{{\left({1 - 0} \right)\left({0 - 4} \right)}}{2} = - 2\end{array}\)
Cách khác:
\(\begin{array}{l}\lim {x_n} = \lim \left( {n - \frac{1}{n}} \right)\left({\frac{{1 - 4n}}{{2{n^2}}}} \right)\\ = \lim \left[ {n\left({1 - \frac{1}{{{n^2}}}} \right).\frac{1}{n}\left({\frac{{1 - 4n}}{{2n}}} \right)} \right]\\ = \lim \left[ {n\left({1 - \frac{1}{{{n^2}}}} \right).\frac{1}{n}\left({\frac{1}{{2n}} - 2} \right)} \right]\\ = \lim \left({1 - \frac{1}{{{n^2}}}} \right)\left({\frac{1}{{2n}} - 2} \right)\\ = \left({1 - 0} \right)\left({0 - 2} \right) = - 2\end{array}\)
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